2013 H2 Mathematics Paper 2 Question 11

Permutations and Combinations (P&C)

Answers

(i)

{\frac{400}{507}\approx0.789}

(ii)

{\frac{4}{9}}

(iii)

{\frac{1201}{6084}\approx0.197}

(iv)

{\frac{1225}{6591}\approx0.186}

Full solutions

(i)

Probability required

\begin{align*} &= \frac{{26 \choose 3}\times 3! \times {9 \choose 2}\times 2!}{26^3 \times 9^2} \\ &= \frac{400}{507} \; \blacksquare \end{align*}

(ii)

Let

{x}

denote the probability that the second digit is higher than the first digit
We note that the probability that the first digit is higher than the second digit is also

{x}

When picking the digits, either the second is higher than the first, or the first is higher than the second, or the two digits are the same
Let

{D}

denote the event that the two digits are the same

\begin{align*} \textrm{P}\left(D\right) &= \frac{{9 \choose 1}}{9^2} \\ &= \frac{1}{9} \end{align*}

\textrm{P}\left(1^{\textrm{st}} > 2^{\textrm{nd}}\right) + \textrm{P}\left(2^{\textrm{nd}} > 1^{\textrm{st}}\right) + \textrm{P}\left(\textrm{digits same}\right) = 1

\begin{align*} x + x + \frac{1}{9} &= 1 \\ 2x &= 1-\frac{1}{9} \\ x &= \frac{4}{9} \; \blacksquare \end{align*}

(iii)

Let

{L}

denote the event that exactly two letters are the same

\begin{align*} \textrm{P}\left(L\right) &= \frac{{26 \choose 2}\times 2!\times \frac{3!}{2!}}{26^3} \\ &= \frac{75}{676} \end{align*}

Note that the events

{D}

and

{L}

independent so

\begin{align*} \textrm{P}\left(L \cap D\right) &= \textrm{P}\left(L\right) \times \textrm{P}\left(D\right) \\ &= \frac{75}{676} \times \frac{1}{9} \\ &= \frac{25}{2028} \end{align*}

The event required is represented by

{(L \cap D') \cup (L' \cap D)}

\begin{align*} & \textrm{P}\Big( (L \cap D') \cup (L' \cap D) \Big) \\ &= \textrm{P}\left(L \cup D\right) - \textrm{P}\left(L \cap D\right) \\ &= \textrm{P}\left(L\right) + \textrm{P}\left(D\right) - 2 \textrm{P}\left(L \cap D\right) \\ &= \frac{75}{676} + \frac{1}{9} - 2\left(\frac{25}{2028}\right) \\ &= \frac{1201}{6084} \; \blacksquare \end{align*}

(iv)

Case 1: Exactly 1 vowel, a repeated consonant and exactly one even digit
Probability of this case

\begin{align*} & = \frac{{5 \choose 1}{21 \choose 1} \frac{3!}{2!}\times{5\choose 1}{4 \choose 1} 2!}{26^3 \times 9^2} \\ & = \frac{175}{19773} \\ \end{align*}

Case 2: Exactly 1 vowel, 2 distinct consonant and exactly one even digit
Probability of this case

\begin{align*} & = \frac{{5 \choose 1}{21 \choose 2}3!\times{5\choose 1}{4 \choose 1} 2!}{26^3 \times 9^2} \\ & = \frac{3500}{19773} \\ \end{align*}

Required probability

\begin{align*} & = \frac{175}{19773} + \frac{3500}{19773} \\ &= \frac{1225}{6591} \; \blacksquare \end{align*}